Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $18.8$ years; the standard deviation is $3.9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $14.9$ years.
Answer: $18.8$ $14.9$ $22.7$ $11$ $26.6$ $7.1$ $30.5$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $18.8$ years. We know the standard deviation is $3.9$ years, so one standard deviation below the mean is $14.9$ years and one standard deviation above the mean is $22.7$ years. Two standard deviations below the mean is $11$ years and two standard deviations above the mean is $26.6$ years. Three standard deviations below the mean is $7.1$ years and three standard deviations above the mean is $30.5$ years. We are interested in the probability of a gorilla living less than $14.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $14.9$ years and the other half $({16\%})$ will live longer than $22.7$ years. The probability of a particular gorilla living less than $14.9$ years is ${16\%}$.